A prolate elliptical dome (fig. 195.) is produced by the revolution of a semi-ellipse round its greater semi-axis.

(523.) The art of turning consists chiefly in the production of surfaces of revolution. The cutting tool imparts the circular form to the body, which is turned by the lathe, and if the cutting tool itself be guided along the lines, by the revolution of which the surface is supposed to be formed, the requisite form will be imparted to the body submitted to the operation. Thus if the cutter be moved along a line parallel to the axis of the lathe, a cylinder will be formed; if it be moved along a straight line, intersecting the axis of the lathe, a cone will be formed; if it be moved along a straight line which is not in the plane of the axis of the lathe, a surface will be formed like that represented in (fig.189.); if it be moved in a semicircle, whose centre lies in the axis of the lathe, a sphere will be formed; and if it be moved in a semi-ellipse, a spheroid will be formed, and

CHAP. XVIII.

OF THE REGULAR SOLIDS.

(524.) A REGULAR solid is a solid all the faces of which are regular polygons, or, rather, regular plane figures; that is, figures which are equiangular and equilateral.

(525.) It is easy to prove that there cannot be more than five regular solids.

1. If the faces be equilateral triangles, solid angles may be formed by their combination in different ways. Three, four, or five angles of 60° may form a solid angle; but if six or more such plane angles were united edge to edge, they would be equal to or greater than 360°, and consequently could not form a solid angle, since the sum of the plane angles forming a solid angle must evidently be less than 360°.

The number of regular solids, therefore, whose faces are equilateral triangles cannot exceed three.

2. If the faces be squares, a solid angle can be formed by three right angles, but not by four, or any greater number, since the sum of four right angles is equal to 360 degrees. There cannot, therefore, be more than one regular solid with square faces.

3. Suppose the faces are regular pentagons. A solid angle may be formed of three angles of a regular pentagon, for the magnitude of the angle of a regular pentagon is six-fifths of a right angle, and therefore the aggregate magnitude of three such angles is eighteenfifths of a right angle, or three right angles and threefifths, which being less than four right angles, a solid angle may therefore be formed by three angles of a regular pentagon. But four or more such angles, being greater than four right angles, cannot form a solid angle. Hence there cannot be more than one regular solid with pentagonal faces.

4. Suppose the faces were regular hexagons. The angles of a regular hexagon are 120 degrees, and three such angles would therefore be equal to 360 degrees. Three angles of a regular hexagon combined would therefore form a plane, and could not form a solid angle; and as four or more such angles would be greater than 360 degrees, they could not form a solid angle.

5. The angles of all regular polygons having more than six sides are greater than one third of four right angles. Consequently three or more such angles combined, amounting to more than 360 degrees, cannot form a solid angle.

Hence no regular solid can have faces with more than five sides. Hence we infer, first, that there cannot be more than five regular solids; secondly, that of these, three have triangular faces, one has a square face, and one a pentagonal face; thirdly, that the solid angles of the three regular solids having triangular faces are formed of three, four, and five plane angles, and that the solid angles of the others are formed of three plane angles.

(526.) To construct a regular solid having triangular faces, whose solid angles shall be composed of three plane angles, let. A B C (fig. 196.) be one of the sides of

such a solid, and let O be the centre of this equilateral triangle, taken upon the perpendicular from the angle A to the side BC, at a distance O a from that side equal to one third

of the length of A a.

From the

point O draw a perpendicular OP to the plane of the triangle ABC.

From the three angles A, B, C, let

lines be inflected on this perpendicular equal to the sides of the equilateral triangle ABC. These lines will meet the perpendicular at the same point P, and

will form the edges of a triangular pyramid, whose faces will be equilateral triangles equal to the base A B C.

To prove that the lines thus inflected will meet the perpendicular OP at the same point, let AP be one of those lines. In the right-angled triangle AOP the square of OP will be equal to the difference between the squares of AP and AO; but the point O being at equal distances from each of the three angles of the triangle ABC, the height of the point at which each of the inflected lines will meet the perpendicular above the point O will be the same, its square being equal to the difference of the squares of the equal inflected lines and the equal distances of the point O from the three angles.

The inclinations of the planes of every pair of faces are equal. Since A a and Pa are both drawn to the middle point of the common base of the equilateral triangles BAC and BPC, they will be perpendicular to that base, consequently, the angle Pa A will be the angle under the planes of the two triangles. For the same reason, PcC will be the angle under the planes of the faces APB and ACB. But since the sides of the triangle A a P, are equal respectively to the sides of the triangle Pc C, the angles of these triangles are equal, and therefore the faces PAB and PBC of the pyramid are equally inclined to the plane of its base.

In the same manner, it may be shown that the planes of all the faces of the solid are equally inclined to each other.

(527.) This regular solid, with four equal and similar triangular faces, is called the regular tetra

edron.

(528.) To determine numerically the volume of a regular tetraedron, whose side is the linear unit. Since A B is the unit, Ba will be, and therefore the square of Ba will be; but the square of A a is the difference between the squares of A B and Ba, and is therefore. But A O being of A a, its square will be of the square of A a, therefore the square of A O is of, or

, or. But since AOP is a right angle, the square of OP is the difference between the squares of A P and A O, that is the difference between 1 and 4, orf. Since then the square of PO is, the line PO itself will

be V

is

2

3

But since the square of A a is, the line A a itself

3

2

,

and this being multiplied into half of BC,

which is, will give.

√3
4

for the area of the triangle ABC. This area being multiplied by of the perpendicular PO, will give the volume of the pyramid.

(529.) The volume of a tetraedron is to that of a cube with an equal edge, therefore as 1 is to six times the square root of 2, or as 1 is to 8.485.

(530.) To construct a regular solid with triangular faces, and whose solid angles are formed by four plane angles.

Construct a square ABCD (fig. 197.), and through its